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12x^2-168x+392=0
a = 12; b = -168; c = +392;
Δ = b2-4ac
Δ = -1682-4·12·392
Δ = 9408
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9408}=\sqrt{3136*3}=\sqrt{3136}*\sqrt{3}=56\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-168)-56\sqrt{3}}{2*12}=\frac{168-56\sqrt{3}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-168)+56\sqrt{3}}{2*12}=\frac{168+56\sqrt{3}}{24} $
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